1720 CHAPTER 54. FUNCTIONAL ANALYSIS APPLICATIONS
If α < β , then(−A)−β (H)⊆ (−A)−α (H) . (54.3.19)
If α ∈ (0,1) , then
(−A)−α =sin(πα)
π
∫∞
0λ−α (λ I−A)−1 dλ (54.3.20)
Proof: Consider 54.3.17.
(−A)−α (−A)−β ≡ 1Γ(α)Γ(β )
∫∞
0
∫∞
0tα−1sβ−1S (t + s)dsdt
Changing variables and using Fubini’s theorem which is justified because of the aboluteconvergence of the iterated integrals, which follows from Corollary 54.3.8, this becomes
1Γ(α)Γ(β )
∫∞
0
∫∞
ttα−1 (u− t)β−1 S (u)dudt
=1
Γ(α)Γ(β )
∫∞
0
∫ u
0tα−1 (u− t)β−1 S (u)dtdu
=1
Γ(α)Γ(β )
∫∞
0S (u)
∫ 1
0(ux)α−1 (u−ux)β−1 udxdu
=1
Γ(α)Γ(β )
(∫ 1
0xα−1 (1− x)β−1 dx
)∫∞
0S (u)uα+β−1du
=1
Γ(α)Γ(β )
(∫ 1
0xα−1 (1− x)β−1 dx
)Γ(α +β )(−A)−(α+β )
= (−A)−(α+β )
This proves the first part of the theorem.Consider 54.3.18. Since A is a closed operator, and approximating the integral with an
appropriate sequence of Riemann sums, (−A) can be taken inside the integral and so
(−A)1
Γ(1)
∫∞
0t1−1S (t)dt =
∫∞
0(−A)S (t)dt
=∫
∞
0− d
dt(S (t))dt = S (0) = I.
Next let x ∈ D(−A) . Then
1Γ(1)
∫∞
0t1−1S (t)dt (−A)x =−
∫∞
0S (t)Axdt
=−∫
∞
0AS (t)xdt =
∫∞
0− d
dt(S (t))dt = Ix
This shows that the integral in which α = 1 deserves to be called A−1 so the definition isnot bad notation. Also, by assumption, A−1 is one to one. Thus
(−A)−1 (−A)−1 x = 0